A small block of mass `1` kg is a circular are of ratius `40` m . The block sides along the track without topping and a frictionnal force acts on it in the direction opposite in the instrmens velocity . The work done in evercoming the friction up to the point `Q` as shown is the figure below is `150 J` (Take the acceleration due to gravity `g = 10 ms^(-2))` The speed of the block when it reaches the point `Q` is
A
(a) `5ms^-1`
B
(b) `10ms^-1`
C
(c) `10sqrt3ms^-1`
D
(d) `20ms^-1`
Text Solution
Verified by Experts
The correct Answer is:
B
Using work energy theorem `mg R sin 30^@+W_f=1/2mv^2` `200-150=v^2/2v=10ms^-1`
A small block of mass 1 kg is a circular are of ratius 40 m . The block sides along the track without topping and a frictionnal force acts on it in the direction opposite in the instrmens velocity . The work done in evercoming the friction up to the point Q as shown is the figure below is 150 J (Take the acceleration due to gravity g = 10 ms^(-2)) The magnitude of the normal reaction that acts on the block at the point Q is
The frictional force acting on 1 kg block is
The acceleration of the 10 kg mass block shown in the figure is : -
As shown in the figure , the friction force acting on the block is
If the mass of block is 1 kg and a force of 10 / sqrt3 N is applied horizontally on the block as shown in the figure . The frictional force acting on the block is :
A block of mass 6 kg is kept on rough surface as shown in figure. Find acceleration and friction force acting on the block. (Take g=10m//s^(2) )
In the figure shown, block of mass 2 kg is very long. Force of friction on 1 kg block is
A block of mass 2kg slipped up a slant plane requires 300J of work. If height of slant is 10m the work done against friction is :
A small block of mass 1 kg is kept on a rough inclined wedge of inclination 45^@ fixed in an elevator. The elevator goes up with a uniform velocity v=2 m//s and the block done not slide on the wedge. Find the work done by the of friction on the block in 1s . (g=10 m//s^(2))
A small cube of mass m slides down a circular path of radius R cut into a large block of mass M. M rests on a table and both blocks move without friction. The blocks initially are at rest and m starts from the top of the path. Find the velocity v of the cube as it leaves the block.
Knowledge Check
A small block of mass 1 kg is a circular are of ratius 40 m . The block sides along the track without topping and a frictionnal force acts on it in the direction opposite in the instrmens velocity . The work done in evercoming the friction up to the point Q as shown is the figure below is 150 J (Take the acceleration due to gravity g = 10 ms^(-2)) The magnitude of the normal reaction that acts on the block at the point Q is
A
` 7.5 N`
B
`8.6 N`
C
11.5 N`
D
`22.5 N`
As shown in the figure , the friction force acting on the block is
A
`50 N`
B
`30 N`
C
`20 N`
D
None
If the mass of block is 1 kg and a force of 10 / sqrt3 N is applied horizontally on the block as shown in the figure . The frictional force acting on the block is :
A
zero
B
`(10)/(sqrt3)`N
C
`(20)/(sqrt3) N`
D
5 N
CENGAGE PHYSICS-WORK, POWER & ENERGY-Archives (linked Comprehension)
